A metallic rod of mass per unit length `0.5 kg m^(-1)` is lying horizontally on a straght inclined plane which makes an angle of `30^(@)` with the horizontal. The rod is not allowed to slide down by flowing a current throguh it when a magnetic field of induction `0.25 T` is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is

Options

A. `7.14 A`
B. `5.98 A`
C. `14.76 A`
D. `11.32 A`

Explanation

To solve the problem step by step, we will analyze the forces acting on the metallic rod and apply the necessary physics principles.

### Step 1: Identify the forces acting on the rod
The rod is subjected to two main forces:
1. The gravitational force (weight) acting downwards, which can be resolved into two components:
- Perpendicular to the inclined plane: \( mg \cos(30^\circ) \)
- Parallel to the inclined plane (down the slope): \( mg \sin(30^\circ) \)

2. The magnetic force acting on the rod due to the current flowing through it in the presence of a magnetic field. The magnetic force can be expressed as:
\[ F_m = I L B \cos(30^\circ) \]
where \( I \) is the current, \( L \) is the length of the rod, and \( B \) is the magnetic field strength.

### Step 2: Set up the force balance equation
Since the rod is not allowed to slide down, the magnetic force must balance the component of the gravitational force acting down the slope:
\[ mg \sin(30^\circ) = I L B \cos(30^\circ) \]

### Step 3: Substitute known values
We know:
- Mass per unit length \( \frac{m}{L} = 0.5 \, \text{kg/m} \)
- Therefore, \( m = 0.5 L \)
- The angle \( \theta = 30^\circ \)
- The magnetic field \( B = 0.25 \, \text{T} \)

Substituting \( m \) in the force balance equation:
\[ (0.5 L) g \sin(30^\circ) = I L B \cos(30^\circ) \]

### Step 4: Simplify the equation
We can cancel \( L \) from both sides (assuming \( L \neq 0 \)):
\[ 0.5 g \sin(30^\circ) = I B \cos(30^\circ) \]

### Step 5: Substitute values for \( g \), \( \sin(30^\circ) \), and \( \cos(30^\circ) \)
Using \( g \approx 10 \, \text{m/s}^2 \):
- \( \sin(30^\circ) = 0.5 \)
- \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \)

Substituting these values into the equation:
\[ 0.5 \cdot 10 \cdot 0.5 = I \cdot 0.25 \cdot \frac{\sqrt{3}}{2} \]

### Step 6: Solve for the current \( I \)
This simplifies to:
\[ 2.5 = I \cdot 0.25 \cdot \frac{\sqrt{3}}{2} \]

Rearranging gives:
\[ I = \frac{2.5}{0.25 \cdot \frac{\sqrt{3}}{2}} = \frac{2.5 \cdot 2}{0.25 \cdot \sqrt{3}} = \frac{5}{0.25 \cdot \sqrt{3}} = \frac{20}{\sqrt{3}} \]

### Step 7: Calculate the numerical value of \( I \)
Calculating \( I \):
\[ I \approx \frac{20}{1.732} \approx 11.55 \, \text{A} \]

### Final Answer
Thus, the current flowing in the rod to keep it stationary is approximately \( 11.55 \, \text{A} \).

---